3
$\begingroup$

In the "Principles of Neural Science" 5th edition in the discussion about electrical synapses it says:

During excitatory synaptic transmission at an electrical synapse, voltage-gated ion channels in the presynaptic cell generate the current that depolarizes the postsynaptic cell. Thus these channels not only depolarize the presynaptic cell above the threshold for an action potential but also generate sufficient ionic current to produce a change in potential in the postsynaptic cell.

To generate such a large current, the presynaptic terminal must be big enough for its membrane to contain many ion channels. At the same time, the postsynaptic cell must be relatively small. This is because a small cell has a higher input resistance (Rin) than a large cell and, according to Ohm's law (V = I x Rin), undergoes a greater voltage change in response to a given presynaptic current (I).

I get that Ohm's law states this, but what is really going on? Why should the resistance encountered by charge as it is moving through gap-junctions from the presynaptic to the postsynaptic cell affect the voltage (electrical potential) across the membrane of the postsynaptic cell with its environment. Once the charge gets to the postsynaptic cell, it seems like it should no longer be affected by the resistance it encountered along the way.

$\endgroup$
1
$\begingroup$

If you consider the electrical synapse a simple series RC circuit, with the resistance in the synapse and the capacitor as the postsynaptic cell, then you are correct that the value of R only changes the charge time for the capacitor (from a DC source, but not it's final voltage (at t = infinity).

enter image description here

($\tau = RC$)

On the other hand, this model is too simplistic for the electrical synapse because you also have losses (discharge) to the outside of the cells. Image from the book:

enter image description here

And a more realistic circuit is created with extra (parallel) resistors below for the extracellular discharge. Note what happens when the 2nd resistor (discharge of postsynaptic cell to inter-cellular space) is introduced in the circuit (switch closed): the voltage charge on capacitor drops significantly. enter image description here

You basically have to "beat" or minimize this voltage drop, and one way to do this is to have a lower series resistor (another is of course a higher isolation [i.e. resistor] from the intercellular space). When the synapse resistance is 10 times lower than in the previous circuit, the voltage drop on the synapse is much lower, and so the voltage on the capacitor is higher, even with the switch closed:

enter image description here

You can play with the circuit here; it's set for the first set of resistor values; you need to right-click on a resistor and pick "edit" to change it's value; I could have added another switch and resistor to achieve that change for the synapse... but I didn't want to complicate the circuit conceptually.

$\endgroup$
  • $\begingroup$ Great detailed answer, but I need a bit more clarification. The book states that the voltage change is greater because: "...a small cell has a higher input resistance (Rin) than a large cell." In your circuit, you seem to have done the opposite, i.e. made the input resistance lower so that more current can flow in to compensate for the current flowing out. $\endgroup$ – learnJ12 Jan 12 '18 at 17:30
  • $\begingroup$ @JayCrosley: yes but a small cell also has an overall higher resistance to the environment/ground. Resistance is inversely proportional with area (of conduction). I'm not sure, but they may overblow the Rin issue because on the subsequent pages (fig 8-4) they don't say that the gap junction depends on cell size. $\endgroup$ – Fizz Jan 12 '18 at 17:47
  • $\begingroup$ Actually by Rin they probably mean the sum of the two resistors (top and right) because that's how Rin is defined. The source/port in this context (of Rin) is the whole presynaptic cell. And a high input impedance/resistance (measured like that) is actually good because it preserves most of the signal. Most electronic amplifiers boast of having high input impedance, precisely for this reason, they don't tax much the source. $\endgroup$ – Fizz Jan 12 '18 at 17:52
  • $\begingroup$ electronics.stackexchange.com/questions/68638/… $\endgroup$ – Fizz Jan 12 '18 at 17:58
  • 1
    $\begingroup$ It's just phrased rather confusingly, because the voltage and resistance in the circuit are being created by the membrane while the current is being created by the gap junction (two different physical substrates). But if we put both the membrane and the gap junction into a circuit, it just looks like a single circuit with a battery (the electrochemical driving force) connected to a wire (the gap junction) connected to a resistor (the membrane). $\endgroup$ – learnJ12 Jan 12 '18 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.